Limits, Continuity & Differentiability
Higher order derivatives and Taylor series
Grade 12
Question:
<p>Given that \(f(x) = x^a\), then \(f(1) - \dfrac{f'(1)}{1!} + \dfrac{f''(1)}{2!} - \dfrac{f'''(1)}{3!} + \cdots + \dfrac{(-1)^n f^n(1)}{n!}\) equals:</p>
<p>\(2^n\)</p>
<p>\(1\)</p>
<p>\(0\)</p>
<p>\((-1)^n\)</p>
Step-by-Step Solution
Key Concept: Recognize this alternating series of derivatives at x=1 as the Taylor series expansion of f(x) = x^a evaluated at x = 1 - 1 = 0. The pattern matches the Taylor series: f(0) = f(1) - f'(1)·1 + f''(1)·1²/2! - ... which is f(1-1).
<p><strong>Step 1:</strong> Recognize the pattern. The given expression is:</p><p>f(1) - f'(1)/1! + f''(1)/2! - f'''(1)/3! + ... + (-1)^n f^n(1)/n!</p><p><strong>Step 2:</strong> This matches the Taylor series expansion formula: f(x) = Σ (-1)^n f^n(a)/n! · (x-a)^n</p><p>Here, with a=1 and evaluating at x=0 (since we have alternating signs with -1 coefficient), we get f(1-1) = f(0).</p><p><strong>Step 3:</strong> For f(x) = x^a, calculate f(0):</p><p>f(0) = 0^a = 0 (for a > 0)</p><p><strong>Step 4:</strong> Alternatively, recognize this as the Taylor expansion: f(1-1) = Σ (-1)^n f^n(1)/n! · 1^n, which directly gives us f(0) = 0^a = 0</p><p>∴ Answer: C (which equals <strong>0</strong>)</p>
Correct Answer: C